4
$\begingroup$

If a sequence diverges to infinity then its subsequences diverge to infinity as well.

How does the fact that the sequence of partial sums $S_n = \sum_{k=1}^{n}\frac{1}{k} \to \infty$ as $n \to \infty$, and the subsequence $S_{n^p} = \sum_{k=1}^{n}\frac{1}{k^p}$ converges when $p>1$ not violate the first proposition? Is my initial assumption wrong? Am I wrong in calling $S_{n^2}$ a subsequence?

Thank you for the help.

$\endgroup$
  • 1
    $\begingroup$ Your $S_{n^p}$ is not a subsequence of $S_n$. For example, $\sum_{k=1}^2 1/k^2 = 1 + 1/4 = 5/4$, whereas the first few terms of the sequence $S_n$ are $1, 3/2, 11/6$. $\endgroup$ – Bungo Dec 1 '17 at 21:36
  • $\begingroup$ $\sum_{k=1}^{n^p}\dfrac1k$ is way bigger than $\sum_{k=1}^n\dfrac1{k^p}$. $\endgroup$ – Jyrki Lahtonen Dec 1 '17 at 21:37
  • 7
    $\begingroup$ You are mixing up sequences and series. $\endgroup$ – Angina Seng Dec 1 '17 at 21:37
  • 1
    $\begingroup$ Put another way, you are mixing up a sequence with the corresponding sequence of its partial sums. 1/n^2 is a sub-sequence of 1/n, but the partial sums of 1/n^2 do not comprise a sub-sequence of the partial sums of 1/n. $\endgroup$ – Sridhar Ramesh Dec 1 '17 at 21:43
8
$\begingroup$

Yes, you are wrong. The series $\sum_{n=1}^\infty\frac1n$ diverges to $+\infty$. What this means is that the sequence$$\tag{1}1,1+\frac12,1+\frac12+\frac13,\ldots$$diverges to $+\infty$. Now consider the series $\sum_{n=1}^\infty\frac1{n^2}$. The first partial sums are$$1,1+\frac14,1+\frac14+\frac19,\ldots$$The terms of this sequence (except for the first one) don't appear in $(1)$. Therefore, it is not a subsequence of $(1)$. So, there is no contradiction.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Note that $S_{n^{p}}=\displaystyle\sum_{k=1}^{n^{p}}\dfrac{1}{k}$, so the $p$-series is not a subsequence of $\{S_{n}\}$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

A little more:

$a_n =1/n$; a subsequence of $ (a_n)_{n \in \mathbb{N}}$ is $b_n =1/n^2$, $(b_n)_{n \in \mathbb{N}}$.

The sequence $a_n$ converges to $0$, so does the subsequence $b_n.$

$S_n = \sum_{k=1}^{n} a_n$ diverges.

$T_m = \sum_{k=1}^{m} b_n$ converges,

But:

$(T_m)_{m \in \mathbb{N}} $ is not a subsequence of $(S_n)_{n \in \mathbb{N}}.$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.